This is a follow-up to this thread:
Proof that sum over autocorrelations is -1/2
I am posting a new thread as that was posted 6 years ago. In that threat the stackexhange author (Kuhlambo) lists some equations from a paper found in https://doi.org/10.1515/ROSE.2009.008, that proof that the sum of the sample autocorrelation coefficients is always equal to -1/2.
I find it hard to understand how the original author of the proof expanded the variance:
$$\sum^{T}_{t=1} (y_t - \overline y)^2$$
into the following expression:
$$\big(\sum^{T}_{t=1} (y_t - \overline y)\big)^2-2 \sum_{h=1}^{T-1}\sum^{T-h}_{t=1} (y_t - \overline y)(y_{t+h} - \overline y)$$
This is a central step to the proof and I cannot see how the author went from one step to the next. The paper does not provide any detail on this derivation, and considers this step as elementary, which it might be given my limited understanding.
Can someone help me understand how we move from one step to the next?
Here is one approach to the calculation. I'll illustrate it when $T=5$. Let's start by making a matrix which I will color in to help the exposition. Let $Y_{t,s} = (y_t - \bar{y})(y_s - \bar{y})$ and define the matrix
\begin{equation} Y = \begin{bmatrix} \color{green}{Y_{1,1}} & \color{blue}{Y_{1,2}} & \color{blue}{Y_{1,3}} & \color{blue}{Y_{1,4}} & \color{blue}{Y_{1,5}} \\ \color{red}{Y_{2,1}} & \color{green}{Y_{2,2}} & \color{blue}{Y_{2,3}} & \color{blue}{Y_{2,4}} & \color{blue}{Y_{25,}} \\ \color{red}{Y_{3,1}} & \color{red}{Y_{3,2}} & \color{green}{Y_{3,3}} & \color{blue}{Y_{3,4}} & \color{blue}{Y_{3,5}} \\ \color{red}{Y_{4,1}} & \color{red}{Y_{4,2}} & \color{red}{Y_{4,3}} & \color{green}{Y_{4,4}} & \color{blue}{Y_{4,5}} \\ \color{red}{Y_{5,1}} & \color{red}{Y_{5,2}} & \color{red}{Y_{5,3}} & \color{red}{Y_{5,4}} & \color{green}{Y_{5,5}} \end{bmatrix} \end{equation}
Now let's a few things about this matrix.
First, $$\sum_{t=1}^T (y_t-\bar{y})^2 = \sum_{t=1}^T (y_t-\bar{y})(y_t -\bar{y}) = \sum_{t=1}^T \color{green}{Y_{t,t}} = \sum_{\color{green}{\text{green}}} \color{green}{Y_{t,t}} $$ the sum of the diagonal $\color{green}{\text{green}}$ terms.
Second \begin{align} \left(\sum_{t=1}^T (y_t - \bar{y})\right)^2 &= \left(\sum_{t=1}^T (y_t - \bar{y})\right)\left(\sum_{s=1}^T (y_s - \bar{y})\right) \\ &=\sum_{s,t=1}^T (y_t - \bar{y})(y_s -\bar{y}) \\ &= \sum_{s,t=1}^T Y_{s,t} \\ &= \sum_{\color{\red}{\text{red}}} \color{red}{Y_{s,t}} + \sum_{\color{blue}{\text{blue}}} \color{blue}{Y_{s,t}} + \sum_{\color{green}{\text{green}}} \color{green}{Y_{s,t}} \end{align} the sum of all of the terms, that is the $\color{red}{\text{red}}$, $\color{blue}{\text{blue}}$ and $\color{green}{\text{green}}$ terms.
The next hard part is to work out what is happening in the sum with $h$ and $t$. Let's fill in a matrix using colors for this.
\begin{equation} \begin{bmatrix} Y_{1,1} & \color{#fa0}{Y_{1,2}} & \color{#cd0}{Y_{1,3}} & \color{#a0f}{Y_{1,4}} & \color{#0af}{Y_{1,5}} \\ Y_{2,1} & Y_{2,2} & \color{#fa0}{Y_{2,3}} & \color{#cd0}{Y_{2,4}} & \color{#a0f}{Y_{2,5}} \\ Y_{3,1} & Y_{3,2} & Y_{3,3} & \color{#fa0}{Y_{3,4}} & \color{#cd0}{Y_{3,5}} \\ Y_{4,1} & Y_{4,2} & Y_{4,3} & Y_{4,4} & \color{#fa0}{Y_{4,5}} \\ Y_{5,1} & Y_{5,2} & Y_{5,3} & Y_{5,4} & Y_{5,5} \end{bmatrix} \end{equation} Each color corresponds to a fixed value of $h$. \begin{align} (\color{#fa0}{h = 1}) \hspace{1cm} \color{#fa0}{Y_{1,2}} + \color{#fa0}{Y_{2,3}} + \color{#fa0}{Y_{3,4}} + \color{#fa0}{Y_{4,5}} &= \sum_{t=1}^{T-1} \color{#fa0}{Y_{t,t+1}} \\ &= \sum_{t=1}^{T-h} \color{#fa0}{Y_{t,t+h}} \\ &= \sum_{t=1}^{T-h}(y_t - \bar{y})(y_{t+h} - \bar{y}) \\ (\color{#cd0}{h = 2}) \hspace{1cm} \color{#cd0}{Y_{1,3}} + \color{#cd0}{Y_{2,4}} + \color{#cd0}{Y_{3,5}} &= \sum_{t=1}^{T-2} \color{#cd0}{Y_{t,t+2}} \\ &= \sum_{t=1}^{T-h} \color{#cd0}{Y_{t,t+h}} \\ &= \sum_{t=1}^{T-2} (y_{t} - \bar{y})(y_{t+h} - \bar{y}) \\ \end{align}
We see that each from $h = 1$ to $T-1$ gives a diagonal band of the matrix $Y$ above the diagonal. Therefore summing over these $h$ gives everything above the main diagonal, and therefore $$\sum_{h=1}^{T-1}\sum_{t=1}^{T-h} (y_t - \bar{y})(y_{t+h} - \bar{y}) = \sum_{\color{blue}{\text{blue}}} \color{blue}{Y_{t,s}}$$ Note that this is not multiplied by 2.
The reason to multiply by 2 is connected to the following fact $$Y_{t,s} = (y_t-\bar{y})(y_s-\bar{y}) = (y_s - \bar{y})(y_t - \bar{y}) = Y_{t,s}$$ From this it follows that $\color{blue}{Y_{t,s}} = \color{red}{Y_{s,t}}$. If we sum over all the $\color{blue}{\text{blue}}$ terms then we obtain $$\sum_{\color{blue}{\text{blue}}} \color{blue}{Y_{t,s}} = \sum_{\color{red}{\text{red}}} \color{red}{Y_{s,t}}.$$ From this we obtain $$\sum_{\color{blue}{\text{blue}}} \color{blue}{Y_{t,s}} + \sum_{\color{red}{\text{red}}} \color{red}{Y_{s,t}} = 2\sum_{\color{blue}{\text{blue}}} \color{blue}{Y_{t,s}} = 2\sum_{h=1}^{T-1}\sum_{t=1}^{T-h} (y_t - \bar{y})(y_{t+h} - \bar{y})$$ Now we can put it all together \begin{align} & \left(\sum_{t=1}^T (y_t - \bar{y})\right)^2 - 2\sum_{h=1}^{T-1}\sum_{t=1}^{T-h} (y_t - \bar{y})(y_{t+h} - \bar{y}) \\ &= \left ( \sum_{\color{\red}{\text{red}}} \color{red}{Y_{s,t}} + \sum_{\color{blue}{\text{blue}}} \color{blue}{Y_{s,t}} + \sum_{\color{green}{\text{green}}} \color{green}{Y_{s,t}} \right ) - \left ( \sum_{\color{blue}{\text{blue}}} \color{blue}{Y_{t,s}} + \sum_{\color{red}{\text{red}}} \color{red}{Y_{s,t}} \right ) \\ &= \sum_{\color{green}{\text{green}}} \color{green}{Y_{s,t}} = \sum_{t=1}^T (y_t-\bar{y})^2 \end{align} where in each line we have substituted in the expressions we derived above.
In short, the equality is really just summing the diagonal of a matrix by summing the entire matrix and then subtracting the off-diagonal entries. Most of the difficulty is in unpacking how this is being done in the summations given.
It's also worth noting that this works for any matrix and we didn't anywhere use that $\bar{y}$ is the mean. It's worth it for your own knowledge to generalize this arbitrary $Y_{t,s}$.